I am confused with the proof of Lemma 1.5.10 in Algebraic Function Fields and Codes by Henning Stichtenoth.
Let $0\ne\omega\in\Omega_F$. Then there is a uniquely determined divisor $W\in M(\omega)$ such that $A≤W$ for all $A\in M(\omega)$.
$\Omega_F$ is the set of Weil differentials of function field $F/K$, and $M(\omega)$ is the set of divisors $A$ such that $\omega$ vanishes on the adele space $\mathcal A_F(A)+F$.
In the proof he said:
... we can choose a divisor $W\in M(\omega)$ of maximal degree. Suppose $W$ does not have the property of our lemma. Then there exists a divisor $A_0\in M(\omega)$ with $A_0\not\le W$, i.e. $v_Q(A_0)>v_Q(W)$ for some divisor $Q$. We claim that $W+Q\in M(\omega)$, which is a contradiction to the maximality of $W$. In fact, consider an adele $\alpha=(\alpha_P)\in\mathcal A_F(W+Q)$. We can write $\alpha=\alpha^\prime+\alpha^{\prime\prime}$ with $$\alpha_P^\prime=\begin{cases}\alpha_P&\text{ for }P\ne Q,\\0&\text{ for }P=Q,\end{cases}\quad\alpha_P^{\prime\prime}=\begin{cases}0&\text{ for }P\ne Q,\\\alpha_Q&\text{ for }P=Q.\end{cases}$$ Then $\alpha^\prime\in\mathcal A_F(W)$ and $\alpha^{\prime\prime}\in\mathcal A_F(A_0)$.
As far as I know, $\alpha^\prime\in\mathcal A_F(W)$ only if $0=v_Q(\alpha_Q^\prime)\ge-v_Q(W)$, but I can see nowhere that $v_Q(W)\ge0$, as well as $v_P(A_0)\ge0$.
Forgive me if this is trivial.
Note that $\alpha'_P$ is the $p$th component of the adele $\alpha'$ and that $\nu_P(0) = \infty$ by definition. Therefore, $\nu_Q(\alpha'_Q) = \nu_Q(0) = \infty > -\nu_Q(W)$ since $\nu_Q(W)$ is necessarily finite. Therefore, you get that $\alpha'\in \mathcal A_F(W)$. This same reasoning also gives you that $\alpha''\in \mathcal A_F(A_0)$.